\(\int \frac {x}{(a+b x^2) \sqrt {c+d x^2}} \, dx\) [705]

Optimal result

Integrand size = 22, antiderivative size = 49 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}} \]

[Out]

-arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(1/2)/(-a*d+b*c)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {455, 65, 214} \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}} \]

[In]

Int[x/((a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

-(ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]]/(Sqrt[b]*Sqrt[b*c - a*d]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d} \\ & = -\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {b} \sqrt {b c-a d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} \sqrt {-b c+a d}} \]

[In]

Integrate[x/((a + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]]/(Sqrt[b]*Sqrt[-(b*c) + a*d])

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {\arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\)	\(38\)
default	\(-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+\frac {2 d \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}-\frac {\ln \left (\frac {-\frac {2 \left (a d -b c \right )}{b}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {d \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-\frac {2 d \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 b \sqrt {-\frac {a d -b c}{b}}}\)	\(300\)

[In]

int(x/(b*x^2+a)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^2+c)^(1/2)/((a*d-b*c)*b)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (39) = 78\).

Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 4.71 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\left [\frac {\log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {b^{2} c - a b d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, \sqrt {b^{2} c - a b d}}, -\frac {\sqrt {-b^{2} c + a b d} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {-b^{2} c + a b d} \sqrt {d x^{2} + c}}{2 \, {\left (b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )}}\right )}{2 \, {\left (b^{2} c - a b d\right )}}\right ] \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/4*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^2 - 4*(b*d*x^2 + 2*b*c -
a*d)*sqrt(b^2*c - a*b*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2))/sqrt(b^2*c - a*b*d), -1/2*sqrt(-b^2*c
+ a*b*d)*arctan(-1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(-b^2*c + a*b*d)*sqrt(d*x^2 + c)/(b^2*c^2 - a*b*c*d + (b^2*c*
d - a*b*d^2)*x^2))/(b^2*c - a*b*d)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (42) = 84\).

Time = 3.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.73 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\begin {cases} \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b \sqrt {\frac {a d - b c}{b}}} & \text {for}\: d \neq 0 \\\begin {cases} \frac {x^{2}}{2 a \sqrt {c}} & \text {for}\: b = 0 \\\tilde {\infty } x^{2} & \text {for}\: \sqrt {c} = 0 \\\frac {\log {\left (2 a \sqrt {c} + 2 b \sqrt {c} x^{2} \right )}}{2 b \sqrt {c}} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]

[In]

integrate(x/(b*x**2+a)/(d*x**2+c)**(1/2),x)

[Out]

Piecewise((atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(b*sqrt((a*d - b*c)/b)), Ne(d, 0)), (Piecewise((x**2/(2*
a*sqrt(c)), Eq(b, 0)), (zoo*x**2, Eq(sqrt(c), 0)), (log(2*a*sqrt(c) + 2*b*sqrt(c)*x**2)/(2*b*sqrt(c)), True)),
True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d}} \]

[In]

integrate(x/(b*x^2+a)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/sqrt(-b^2*c + a*b*d)

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx=\frac {\mathrm {atan}\left (\frac {b\,\sqrt {d\,x^2+c}}{\sqrt {a\,b\,d-b^2\,c}}\right )}{\sqrt {a\,b\,d-b^2\,c}} \]

[In]

int(x/((a + b*x^2)*(c + d*x^2)^(1/2)),x)

[Out]

atan((b*(c + d*x^2)^(1/2))/(a*b*d - b^2*c)^(1/2))/(a*b*d - b^2*c)^(1/2)